A girl standing on a bridge throws a stone vertically downward with an intial velocity of 15.0 m/s into the river below. If the stone hit's the water 2.00 seconds later what is the height of the bridge above the water?

Answer 1

h=vi*t+1/2 a t^2
h=-15*2-4.9 *4
h=-30-19.6=-49.6m to the water, so the height of the bridge is 49.6 above the water.

Answer 2

Everything up is positive
Everything Down is negaitive

velocity initial = -15.0m/s
distance final = 0m
distance initial = ?
time = 2s

df=v * t+di
0 = -15 * (2)+di
di = 30 m
therefor the bridge is 30m ablive the water.

Not sure if this is entirely correct.

Answer 3

physics Kid: Nope, it is not correct.